Proof of Closed Subset of a Compact Space is Compact

By Compact Subset Criterion via Open Covers in the Ambient Space it suffices to show that for every open cover $(U_i)_{i\in I}$ of $A$ in $X$ there is a finite subset $J\subseteq I$ with $A\subseteq\bigcup_{j\in J}U_j$.

So let $(U_i)_{i\in I}$ be an open cover of $A$ in $X$.

Case $I=\emptyset$. Then $A\subseteq\bigcup_{i\in\emptyset}U_i=\emptyset$, so $A=\emptyset$, and $J=\emptyset$ is a finite subset of $I$ with $A\subseteq\bigcup_{j\in J}U_j=\emptyset$.

Case $I\ne\emptyset$. Since $A$ is closed in $X$, the complement $X\setminus A$ belongs to $\mathcal{T}$. For $i\in I$ put

Each $W_i$ is open: the set $P_i=\{U_i,\ X\setminus A\}$ is a set whose elements are subsets of $X$, and the family of subsets of $X$ $(G_p)_{p\in P_i}$ defined by $G_p=p$ consists of members of $\mathcal{T}$ and has union $\bigcup_{p\in P_i}G_p=U_i\cup(X\setminus A)=W_i$, so $W_i\in\mathcal{T}$ by condition 2 of Topological Space.

The family $(W_i)_{i\in I}$ covers $X$. Indeed, let $x\in X$. If $x\in A$, then $x\in U_i$ for some $i\in I$, because $(U_i)_{i\in I}$ covers $A$, and hence $x\in W_i$. If $x\notin A$, then $x\in X\setminus A$, and since $I\ne\emptyset$ there is some $i\in I$, for which $x\in W_i$. Therefore $X\subseteq\bigcup_{i\in I}W_i$.

Since $X$ is compact, there is a finite subset $J\subseteq I$ with $X\subseteq\bigcup_{j\in J}W_j$.

Finally, $A\subseteq\bigcup_{j\in J}U_j$. Indeed, let $x\in A$. Then $x\in X$, so $x\in W_j$ for some $j\in J$, that is, $x\in U_j$ or $x\in X\setminus A$. The second alternative is impossible because $x\in A$, so $x\in U_j$ with $j\in J$.

In both cases the required finite subset $J$ exists, so $A$ is compact in $X$.